I have a question regarding an alternate form of the Bessel equation and how that alternate form translates to the modified Bessel equation and its solution. The modified form is from: http://mathworld.wolfram.com/BesselDifferentialEquation.html

and looks like this: $$ \frac{d^2 y}{dx^2}+\frac{1-2\alpha}{x}\frac{dy}{dx}+\left(\beta^2\gamma^2x^{2\gamma-2}+\frac{\alpha^2-n^2\gamma^2}{x^2}\right)y=0 $$ and has the following solutions: $$ y= \begin{cases} x^\alpha\left[AJ_n(\beta x^\gamma)+BY_n(\beta x^\gamma)\right] &\text{ for integer }n \\ \\ x^\alpha\left[AJ_n(\beta x^\gamma)+BJ_{-n}(\beta x^\gamma)\right] &\text{ for noninteger }n \end{cases} $$ For the moment, I am only concerned with the case where $\gamma=1$ and $\alpha=n$ so the equation simplifies to: $$ \frac{d^2 y}{dx^2}+\frac{1-2\alpha}{x}\frac{dy}{dx}+\beta^2y=0 $$ with the following solutions: $$ y= \begin{cases} x^\alpha\left[AJ_\alpha(\beta x)+BY_\alpha(\beta x)\right] &\text{ for integer }\alpha \\ \\ x^\alpha\left[AJ_\alpha(\beta x)+BJ_{-\alpha}(\beta x)\right] &\text{ for noninteger }\alpha \end{cases} $$ What I'd like to know is if instead of being the unmodified Bessel equation, the equation was in the form of the modified Bessel equation as shown below, what would the solutions be? $$ \frac{d^2 y}{dx^2}+\frac{1-2\alpha}{x}\frac{dy}{dx}-\beta^2y=0 $$ A quick search on wolframalpha tells me they might look like this: $$ y= \begin{cases} x^\alpha\left[AJ_\alpha(-i\beta x)+BY_\alpha(-i\beta x)\right] &\text{ for integer }\alpha \\ \\ x^\alpha\left[AJ_\alpha(-i\beta x)+BJ_{-\alpha}(-i\beta x)\right] &\text{ for noninteger }\alpha \end{cases} $$ Could these then be translated to the modified Bessel functions so: $$ y= \begin{cases} x^\alpha\left[CI_\alpha(\beta x)+DK_\alpha(\beta x)\right] &\text{ for integer }\alpha \\ \\ x^\alpha\left[CI_\alpha(\beta x)+DI_{-\alpha}(\beta x)\right] &\text{ for noninteger }\alpha \end{cases} $$

Any insights you could provide would be greatly appreciated, thanks!

  • $\begingroup$ Yes, since the modified Bessel functions are more or less regular Bessel functions on the imaginary line $\endgroup$
    – Dylan
    Jan 23, 2019 at 4:11
  • $\begingroup$ Yes, except that you shouldn't keep using both $\alpha$ and $n$ if they are assumed to be equal. $\endgroup$
    – Christoph
    Jan 23, 2019 at 5:35
  • $\begingroup$ Posted also on MathOverflow: Alternate forms of the Bessel equation. I think that this answer contains very reasonable advice about cross-posting. The most important thing is probably to link both copies to each other. (Of course, you can have a look at other discussions about cross-posting, too.) $\endgroup$ Jan 23, 2019 at 7:33
  • $\begingroup$ edited so only $\alpha$ is used in simplified versions $\endgroup$
    – Travis
    Jan 23, 2019 at 14:16

1 Answer 1


Yes. If we start from \begin{equation} \frac{d^2y}{dx^2} + \frac{1-2\alpha}{x} \frac{dy}{dx} - \beta^2 y = 0, \end{equation} we write $\beta^{\alpha} y = \xi^{\alpha} z(\xi)$ with a new independent variable $\xi = \beta x$ and with a new dependent variable $z$. Computing the first two derivatives of $y$ by the product and chain rules yields \begin{eqnarray} \beta^{\alpha} \frac{dy}{dx} &=& \alpha \xi^{\alpha-1} \beta z + \xi^{\alpha} \beta \frac{dz}{d\xi},\\ \beta^{\alpha} \frac{d^2y}{dx^2} &=& \alpha(\alpha-1) \xi^{\alpha-2} \beta^2 z + 2 \alpha \xi^{\alpha-1} \beta^2 \frac{dz}{d\xi} + \xi^{\alpha} \beta^2 \frac{d^2z}{d\xi^2}. \end{eqnarray} We now obtain \begin{eqnarray} 0 &=& \beta^{\alpha} \left( \frac{d^2y}{dx^2} + \frac{1-2\alpha}{x} \frac{dy}{dx} - \beta^2 y \right) = \beta^{\alpha} \frac{d^2y}{dx^2} + \frac{1-2\alpha}{\xi} \beta \beta^{\alpha} \frac{dy}{dx} - \beta^2 \beta^{\alpha} y\\ &=& \dots = \xi^{\alpha-2} \beta^2 \left( \xi^2 \frac{d^2z}{d\xi^2} + \xi \frac{dz}{d\xi} - (\xi^2 + \alpha^2) z \right). \end{eqnarray} So now we have the modified Bessel's equation $\xi^2 \frac{d^2z}{d\xi^2} + \xi \frac{dz}{d\xi} - (\xi^2 + \alpha^2) z = 0$ for $z$, and a fundamental system of solutions is given by $\{I_{\alpha}(\xi),K_{\alpha}(\xi)\}$ for any value of $\alpha$, or by $\{I_{\alpha}(\xi),I_{-\alpha}(\xi)\}$, if $\alpha \not \in \mathbb{Z}$.

Transforming back we obtain $y(x) = \beta^{-\alpha}(\beta x)^{\alpha} z(\beta x) = x^{\alpha} z(\beta x)$.

The only thing that I don't like about the Wolfram notation is that it might be misinterpreted in the way that $\{I_{\alpha}(\xi),K_{\alpha}(\xi)\}$ is a fundamental system of solutions only if $\alpha \in \mathbb{Z}$, which is not true.

  • $\begingroup$ Thank you for the help and also that last clarification. That's exactly how I did interpret the Wolfram notation. I'm am concerned with any value for $\alpha$, not necessarily when it is an integer. As such, I assume I should stick with using $I_\alpha$ and $K_\alpha$ for my system of solutions? $\endgroup$
    – Travis
    Jan 23, 2019 at 14:23

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